What is the Nernst equation and how is it used?

The Nernst equation gives the cell potential at non-standard conditions: E = E° − (RT/nF)ln Q, or at 25°C: E = E° − (0.0592/n)log Q. It shows how concentration affects EMF. At equilibrium, E = 0 and log K = nE°/0.0592.

State Faraday's first and second laws of electrolysis.

First law: Mass deposited is directly proportional to charge passed. m = ZQ = ZIt. Second law: When the same charge passes through different electrolytes, masses deposited are proportional to their equivalent weights.

Tamil Nadu Class 12 Chemistry — Electrochemistry Notes, Formulae & Important Questions

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Quick Summary

Electrochemistry connects chemical reactions with electrical energy. The chapter has three big zones: conductance of electrolytes, electrochemical cells and EMF, and applications such as batteries, fuel cells, electrolysis, and corrosion. Board exams consistently pick numericals from the Nernst equation and Faraday's laws, so those two areas deserve special attention.

1. Conductance Basics

Electrical conductance in solutions arises from the movement of ions. Unlike metallic conductors, ionic conductors carry charge through ion migration rather than electron flow.

Quantity	Symbol	Meaning	Unit
Resistance	R	Opposition to current	Ohm (Ω)
Conductance	G	Reciprocal of resistance, G = 1/R	Siemens (S)
Specific conductance	κ (kappa)	Conductance of 1 m cube of solution	S m⁻¹
Molar conductivity	Λm	Conductance of volume containing 1 mol	S m² mol⁻¹

Key relations:

R = ρ (l/A)     → ρ is resistivity (Ω m)
κ = 1/ρ
Λm = κ × 1000 / C    (C in mol L⁻¹, Λm in S cm² mol⁻¹)

Variation with dilution:

Specific conductance κ decreases on dilution — fewer ions per unit volume.
Molar conductivity Λm increases on dilution — at lower concentrations, ion–ion interactions decrease and mobility improves.

2. Kohlrausch's Law of Independent Migration of Ions

At infinite dilution, every ion contributes independently to the molar conductivity of the electrolyte.

Λm° = λ+° + λ-°

For an electrolyte AmBn:

Λm° = m·λ+° + n·λ-°

Applications of Kohlrausch's Law:

1. Weak electrolytes: Their Λm° cannot be measured directly (curve never levels off). It is calculated using strong-electrolyte data.

Example: Λm°(CH₃COOH) = Λm°(CH₃COONa) + Λm°(HCl) − Λm°(NaCl)

2. Degree of dissociation of weak electrolytes:

   α = Λm(at concentration C) / Λm°

3. Dissociation constant Ka:

   Ka = Cα² / (1 − α)

4. Solubility of sparingly soluble salts: Measure κ of saturated solution → calculate Λm → find concentration (solubility).

3. Electrochemical Cells

Galvanic (Voltaic) Cell

A galvanic cell converts chemical energy to electrical energy through a spontaneous redox reaction.

Daniell Cell:

Zn | Zn²⁺(aq) || Cu²⁺(aq) | Cu
  (anode)            (cathode)

Anode (oxidation): Zn → Zn²⁺ + 2e⁻ (negative electrode)
Cathode (reduction): Cu²⁺ + 2e⁻ → Cu (positive electrode)
Net: Zn + Cu²⁺ → Zn²⁺ + Cu

The salt bridge (KCl or KNO₃ in agar gel) completes the circuit by allowing ion flow while preventing mixing of solutions, thus maintaining electrical neutrality in both half-cells.

Cell EMF:

Ecell = Ecathode° − Eanode°

Standard Electrode Potential

Measured against the Standard Hydrogen Electrode (SHE), which is arbitrarily assigned E° = 0.00 V.

SHE conditions: H₂ gas at 1 atm, H⁺ at 1 M, 25°C.

The electrochemical series lists E° values from most negative (strongest reducing agents) to most positive (strongest oxidising agents). A metal higher in the series can displace metals below it from their salt solutions.

4. Nernst Equation

At non-standard conditions:

Ecell = Ecell° − (RT / nF) ln Q

At 298 K (substituting R = 8.314 J mol⁻¹ K⁻¹, F = 96500 C mol⁻¹):

Ecell = Ecell° − (0.0592 / n) log Q

For the Daniell cell:

Ecell = Ecell° − (0.0592/2) log ([Zn²⁺]/[Cu²⁺])

At equilibrium: Ecell = 0

log K = n·Ecell° / 0.0592

Relation Between EMF and Gibbs Energy

ΔG = −nFEcell
ΔG° = −nFEcell°

If Ecell > 0, then ΔG < 0 → reaction is spontaneous. This is the fundamental link between electrochemistry and thermodynamics.

5. Faraday's Laws of Electrolysis

First Law

The mass of substance deposited at an electrode is directly proportional to the quantity of charge passed.

m = Z × Q = Z × I × t

where Z is the electrochemical equivalent (g C⁻¹), I is current (A), t is time (s).

Z = M / (n × F) = Equivalent weight / 96500

Second Law

When the same quantity of charge is passed through different electrolytes connected in series, the masses deposited are proportional to their equivalent weights.

m₁/m₂ = Eq.wt₁ / Eq.wt₂

6. Worked Numericals

Problem 1 — Standard EMF of Daniell Cell

GivenE°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V

Find: Standard cell EMF.

Solution

Ecell° = Ecathode° − Eanode°
       = +0.34 − (−0.76)
       = +0.34 + 0.76
       = 1.10 V

Since Ecell° > 0, the reaction is spontaneous.

ΔG° = −nFEcell° = −2 × 96500 × 1.10 = −212,300 J = −212.3 kJ

Problem 2 — Nernst Equation Application

GivenDaniell cell with [Zn²⁺] = 0.10 M, [Cu²⁺] = 0.01 M, Ecell° = 1.10 V, n = 2

Find: Ecell at these concentrations.

Solution

Ecell = Ecell° − (0.0592/n) log([Zn²⁺]/[Cu²⁺])

     = 1.10 − (0.0592/2) log(0.10/0.01)

     = 1.10 − (0.0296) log(10)

     = 1.10 − (0.0296)(1)

     = 1.10 − 0.0296

     = 1.0704 V ≈ 1.07 V

When [Zn²⁺] increases relative to [Cu²⁺], EMF decreases (reaction favours forward reaction less at these concentrations).

Problem 3 — Faraday's First Law (Copper Deposition)

GivenCurrent = 2 A, time = 965 s, electrolyte = CuSO₄, atomic mass of Cu = 64 g mol⁻¹, n = 2

Find: Mass of copper deposited.

Solution

Step 1: Charge passed
Q = I × t = 2 × 965 = 1930 C

Step 2: Electrochemical equivalent of Cu
Eq. wt of Cu = 64/2 = 32 g/equivalent
Z = 32 / 96500 = 3.316 × 10⁻⁴ g C⁻¹

Step 3: Mass deposited
m = Z × Q = 3.316 × 10⁻⁴ × 1930
  = 0.64 g

Answer0.64 g of copper is deposited.

Problem 4 — Molar Conductivity Calculation

GivenSpecific conductance of a 0.02 M KCl solution = 2.768 × 10⁻³ S cm⁻¹

Find: Molar conductivity.

Solution

Λm = (κ × 1000) / C
   = (2.768 × 10⁻³ × 1000) / 0.02
   = 2.768 / 0.02
   = 138.4 S cm² mol⁻¹

7. Applications

Batteries

Type	Example	Key Feature
Primary	Dry cell (Leclanche), Mercury cell	Non-rechargeable
Secondary	Lead storage, Ni-Cd, Li-ion	Rechargeable
Fuel cell	H₂–O₂ fuel cell	Continuous fuel supply, product = H₂O

Lead storage battery: 12 V, used in cars. During discharge: Pb anode, PbO₂ cathode, H₂SO₄ electrolyte. Charging reverses both reactions.

Electrolysis Examples

Water: 2H₂O → 2H₂ + O₂ (H₂ at cathode, O₂ at anode)
Brine (Chlor-alkali process): NaCl + H₂O → NaOH + Cl₂ + H₂. Products: Cl₂ at anode, H₂ at cathode, NaOH in solution.
Electroplating: Object to be plated = cathode; metal to be deposited = anode; salt of that metal = electrolyte.

Corrosion

Iron rusting is an electrochemical process: Fe acts as anode (oxidised to Fe²⁺), moist area acts as cathode (O₂ reduced). Prevention: painting, galvanising (zinc coating), sacrificial anode, cathodic protection.

8. Important Board Questions with Model Answers

One-Mark Questions

Q1. Define specific conductance. The conductance of a conductor of unit length and unit cross-sectional area. SI unit: S m⁻¹.

Q2. Write the cell notation for the Daniell cell.

Answer

Zn	Zn²⁺(aq)		Cu²⁺(aq)	Cu

Q3. What is the function of a salt bridge? It maintains electrical neutrality in both half-cells by allowing free movement of ions between the two solutions without mixing them.

Q4. State Kohlrausch's law. At infinite dilution, each ion contributes independently and additively to the molar conductivity: Λm° = λ+° + λ-°.

Q5. Write the relation between ΔG and Ecell. ΔG = −nFEcell

Short-Answer Questions (2–3 marks)

Q1. Distinguish galvanic and electrolytic cells.

Answer

Feature	Galvanic Cell	Electrolytic Cell
Energy conversion	Chemical → Electrical	Electrical → Chemical
Reaction	Spontaneous (ΔG < 0)	Non-spontaneous (ΔG > 0)
Anode	Negative (oxidation)	Positive (oxidation)
Cathode	Positive (reduction)	Negative (reduction)
Example	Daniell cell	Electroplating

Q2. Derive the Nernst equation for the Daniell cell and explain how to find K from it.

Answer

For Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2:

Ecell = 1.10 − (0.0592/2) log([Zn²⁺]/[Cu²⁺])

At equilibrium, Ecell = 0:

0 = 1.10 − (0.0592/2) log K
log K = (2 × 1.10) / 0.0592 = 37.16
K = 10^37.16 ≈ 1.45 × 10³⁷

The enormously large K confirms the reaction goes essentially to completion.

Q3. State and explain Faraday's Second Law with an example.

Answer

When the same quantity of electricity passes through solutions of AgNO₃ and CuSO₄ in series:

m(Ag) / m(Cu) = Eq.wt(Ag) / Eq.wt(Cu)
              = 108 / 32 = 3.375

So 3.375 g of silver deposits for every 1 g of copper deposited.

Long-Answer Questions (5 marks)

Q1. Derive the relation between EMF, Gibbs energy, and equilibrium constant.

Answer

For a reversible cell doing electrical work: 1. Work done by cell = charge × EMF = nF × Ecell 2. At constant T and P, maximum electrical work = −ΔG 3. Therefore: ΔG = −nFEcell 4. Under standard conditions: ΔG° = −nFEcell° 5. Also, ΔG° = −RT ln K → −nFEcell° = −RT ln K 6. Therefore: Ecell° = (RT/nF) ln K = (0.0592/n) log K at 298 K

This links three quantities: spontaneity (ΔG), cell potential (Ecell), and extent of reaction (K).

Q2. Explain molar conductivity, its variation with concentration, and Kohlrausch's law with applications.

Answer

Molar conductivity Λm = κ × 1000/C increases with dilution because:

At high concentrations, strong interionic attractions reduce ionic mobility.
Dilution reduces these interactions; for strong electrolytes Λm increases linearly with √c.
For weak electrolytes, Λm increases sharply at very low concentrations due to increased dissociation.

Kohlrausch's law: Λm° = λ+° + λ-° (each ion contributes independently).

Application: Λm°(CH₃COOH) = Λm°(CH₃COONa) + Λm°(HCl) − Λm°(NaCl) = 91 + 426.2 − 126.4 = 390.8 S cm² mol⁻¹

Degree of dissociation α = Λm / Λm°. This gives Ka = Cα² / (1 − α).

Common Mistakes

Writing Ecell = Eanode − Ecathode (correct is Ecathode − Eanode)
Forgetting the value of n (electrons transferred) in Nernst and Faraday calculations
Confusing specific conductance with molar conductivity when dilution effects are discussed
Using log instead of ln (or vice versa) without appropriate correction factor (2.303)
Treating corrosion as a purely physical phenomenon rather than electrochemical

Revision Checklist

[ ] Conductance formulas and units
[ ] Kohlrausch law and all four applications
[ ] Daniell cell electrode reactions, salt bridge function, cell notation
[ ] Nernst equation — derivation concept and three numericals
[ ] Relations among Ecell, ΔG, and K with numerical application
[ ] Faraday's First Law numerical (2–3 problems)
[ ] Faraday's Second Law with equivalent weight comparison
[ ] Batteries — one primary, one secondary, one fuel cell
[ ] Chlor-alkali electrolysis products
[ ] Corrosion mechanism and two prevention methods

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Frequently Asked Questions

A galvanic cell converts chemical energy to electrical energy (spontaneous reaction, ΔG < 0). An electrolytic cell converts electrical energy to chemical energy (non-spontaneous, requires external power). Examples: galvanic — Daniel cell, dry cell; electrolytic — electroplating, electrolysis of water.